3.604 \(\int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\)

Optimal. Leaf size=110 \[ \frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}+\frac {2 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b}+\frac {7 c \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}} \]

[Out]

1/5*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b/c+7/15*c*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+2/15*(-
c+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b

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Rubi [A]  time = 0.28, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4397, 3800, 4001, 3792} \[ \frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}+\frac {2 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b}+\frac {7 c \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2*(a + b*x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

(7*c*Tan[2*a + 2*b*x])/(15*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (2*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x
])/(15*b) + ((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b*c)

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \sec ^3(2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac {(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}+\frac {2 \int \sec (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \left (\frac {3 c}{2}+c \sec (2 a+2 b x)\right ) \, dx}{5 c}\\ &=\frac {2 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b}+\frac {(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}+\frac {7}{15} \int \sec (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac {7 c \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {2 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b}+\frac {(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 62, normalized size = 0.56 \[ \frac {(5 \cos (a+b x)+2 \cos (5 (a+b x))) \csc (a+b x) \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2*(a + b*x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

((5*Cos[a + b*x] + 2*Cos[5*(a + b*x)])*Csc[a + b*x]*Sec[2*(a + b*x)]^2*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/
(15*b)

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fricas [A]  time = 1.50, size = 84, normalized size = 0.76 \[ \frac {\sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 10 \, \tan \left (b x + a\right )^{2} + 7\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{15 \, {\left (b \tan \left (b x + a\right )^{5} - 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*(15*tan(b*x + a)^4 - 10*tan(b*x + a)^2 + 7)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b
*x + a)^5 - 2*b*tan(b*x + a)^3 + b*tan(b*x + a))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.17, size = 88, normalized size = 0.80 \[ \frac {\sqrt {2}\, \sqrt {\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \cos \left (b x +a \right ) \left (32 \left (\cos ^{4}\left (b x +a \right )\right )-40 \left (\cos ^{2}\left (b x +a \right )\right )+15\right ) \sqrt {4}}{30 b \sin \left (b x +a \right ) \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

1/30*2^(1/2)/b*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*cos(b*x+a)*(32*cos(b*x+a)^4-40*cos(b*x+a)^2+15)/sin(b
*x+a)/(2*cos(b*x+a)^2-1)^2*4^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 11.03, size = 148, normalized size = 1.35 \[ \frac {4\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,10{}\mathrm {i}+b\,x\,10{}\mathrm {i}}\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{15\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)/cos(2*a + 2*b*x)^3,x)

[Out]

(4*(exp(a*4i + b*x*4i)*5i + exp(a*6i + b*x*6i)*5i + exp(a*10i + b*x*10i)*2i + 2i)*((c*(exp(a*2i + b*x*2i)*1i -
1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/(15*b*(exp(a*2i
+ b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)**3*(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)

[Out]

Timed out

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